Tor Flatness and Global Dimension

For $r\in R$, $d((pr)\otimes q)=d_{P}pr\otimes q+(-1{)}^{p}pr\otimes d_{Q}q=d(p\otimes rq)$, since both differentials are $R$-linear; thus the formula descends from elementary tensors.

Applying $d$ again gives $d_P^{2}p\otimes q+(-1{)}^{p-1}{d}_{P}p\otimes d_{Q}q+(-1{)}^p{d}_{P}p\otimes d_{Q}q+p\otimes d_Q^{2}q$.

The first and last terms vanish and the two middle terms cancel. Linearity then proves $d^2=0$ on every finite sum in each total degree.

The augmented complex $P_1\to P_0\to M\to 0$ remains right exact after $N{\otimes }_R-$, so $H_0(N{\otimes }_R{P}_{\bullet })=\mathrm{coker}(N\otimes P_1\to N\otimes P_0)$.

That cokernel is $N{\otimes }_{R}M$ by the displayed right-exact sequence, so the left-resolved construction has the asserted degree-zero value.

The same calculation for $Q_{\bullet }{\otimes }_{R}M$ gives $\mathrm{coker}(Q_1\otimes M\to Q_0\otimes M)=N{\otimes }_{R}M$, proving both identifications.

A comparison map lifting $v$ is a chain map $P_{\bullet }\to P_{\bullet }^{\mathrm{\prime }}$; tensoring with $N$ gives a chain map $N\otimes P_{\bullet }\to N\otimes P_{\bullet }^{\mathrm{\prime }}$, while $u\otimes 1$ handles the first variable.

Chain-homotopic comparison maps induce the same map on homology, so the map of $\mathrm{Tor}$ does not depend on the chosen lift of $v$.

Composition of comparison maps is chain-homotopic to a comparison map for the composite, hence identity and composition laws hold on homology; the right-resolved construction is identical with the variables interchanged.

Use the length-zero projective resolution $0\to M\stackrel{1}{\to }M\to 0$, concentrated in degree $0$.

After tensoring with $N$, the resulting complex is concentrated in degree $0$ with term $N{\otimes }_{R}M$.

Its homology in positive degrees is zero. Resolving a projective right module instead gives the symmetric conclusion.

Choose a free module $F$ and a module $P^{\mathrm{\prime }}$ with $F\cong P\oplus P^{\mathrm{\prime }}$.

For every exact sequence of right modules, tensoring with $F$ is a direct sum of copies of that sequence and is exact; tensoring with $P$ is a direct summand of this exact complex.

A direct summand of an exact complex is exact, so $-{\otimes }_{R}P$ is exact and $P$ is flat. The same direct-summand argument proves the right-handed statement.

The module $Q_p$ is flat as a right module, so $Q_p{\otimes }_R-$ preserves the exact augmented complex $P_{\bullet }\to M$.

Its degree-$q$ terms are exactly $Q_p{\otimes }_R{P}_q$, and its augmentation is $Q_p{\otimes }_R{P}_0\to Q_p{\otimes }_{R}M$.

Thus the row indexed by $p$ is exact, including the augmentation; this is the claimed augmented-row condition.

The module $P_q$ is flat as a left module, so $-{\otimes }_R{P}_q$ preserves exactness of $Q_{\bullet }\to N$.

The resulting degree-$p$ terms are $Q_p{\otimes }_R{P}_q$, with augmentation $Q_0{\otimes }_R{P}_q\to N{\otimes }_R{P}_q$.

Hence every augmented fixed-$q$ row is exact, on the correct right-module/left-module tensor convention.

The augmentations give degreewise surjective chain maps $a:\mathrm{Tot}K\to N{\otimes }_R{P}_{\bullet }$ and $b:\mathrm{Tot}K\to Q_{\bullet }{\otimes }_{R}M$, supported respectively on $p=0$ and $q=0$. Their surjectivity and exact augmented fixed-$q$ and fixed-$p$ complexes follow from The augmented fixed-q rows of the tensor double complex are exact and The augmented rows of the tensor double complex are exact, respectively. The differential is $d_h+d_v$, with $d_v=(-1{)}^p(1\otimes d_P)$ as in The first-quadrant tensor double complex of two projective resolutions.

The kernel of $a$ is the total of the double complex obtained by replacing $K_{0,q}$ by $\ker (Q_0\otimes P_q\to N\otimes P_q)$. Every fixed-$q$ horizontal complex is exact. For a total cycle $z$ of degree $n\ge 0$, take its largest nonzero $q$ component $z_{p,q}$. The cycle equation at that $q$ says $d_h{z}_{p,q}=0$, because the next higher $q$ component is zero. Horizontal exactness gives $y_{p+1,q}$ with $d_{h}y=z_{p,q}$. Subtracting $dy$ removes that component and introduces only a component at $q-1$. Repeating terminates at $q=0$ and expresses $z$ as a boundary. Thus $\ker a$ is acyclic, including degree zero; negative degrees are zero. This is the filtration by $q$, not by $p$.

The kernel of $b$ is obtained by replacing $K_{p,0}$ by $\ker (Q_p\otimes P_0\to Q_p\otimes M)$. Its fixed-$p$ vertical complexes are exact, with multiplication of their differentials by $(-1{)}^p$ harmless. Now eliminate a cycle's largest $p$ component by solving $d_{v}y=z_{p,q}$ in bidegree $(p,q+1)$. Subtracting $dy$ leaves only lower $p$ components; finitely many repetitions show that $\ker b$ is acyclic. This uses the filtration by $p$, not by $q$.

The short exact sequences of each kernel, total complex and edge, together with The long exact sequence in homology, make both $a$ and $b$ quasi-isomorphisms. Hence $H_i(N{\otimes }_R{P}_{\bullet })\stackrel{H_i(a)}{\leftarrow }{H}_i(\mathrm{Tot}K)\stackrel{H_i(b)}{\to }{H}_i(Q_{\bullet }{\otimes }_{R}M)$ gives the asserted isomorphism $H_i(b)H_i(a{)}^{-1}$.

Under DC, maps of modules lift to maps of their supplied resolutions by Projective comparison maps exist. The resulting tensor double-complex maps commute with $a,b$, proving naturality of the ratio in step 3.1. Two lifts are chain-homotopic by Projective comparison maps are unique up to chain homotopy. Tensor homotopies in the first factor are $h_Q\otimes 1$ and in the second factor $(-1{)}^{p}1\otimes h_P$; direct substitution gives the total homotopy equation. Thus induced homology maps are independent of lifts. Taking module identities also gives independence and coherence under change of the supplied resolutions.

Each comparison map gives a morphism between the two tensor double complexes, compatible with both augmented edges.

The two edge-to-total quasi-isomorphisms therefore form a commutative square on homology, so the balance isomorphism commutes with the comparison maps.

Comparison maps are unique up to chain homotopy, and homotopic maps induce the same homology map; successive changes compose coherently.

By The horseshoe lemma for projective resolutions, choose a projective horseshoe resolution of the middle module that fits with resolutions of the outer modules into a degreewise split short exact sequence. Tensoring it with $N$ preserves the degreewise splittings, hence gives a short exact sequence of chain complexes.

The homology long-exact-sequence construction supplies the displayed connecting maps and exactness.

Change-of-resolution coherence identifies the three homology families with the fixed balanced Tor functor and makes the sequence natural. Its degree-zero tail is $N{\otimes }_R{M}^{\mathrm{\prime }}\to N{\otimes }_{R}M\to N{\otimes }_R{M}^{\mathrm{\prime }\mathrm{\prime }}\to 0$, so no unclaimed left exactness is introduced.

Resolving the right modules gives a short exact sequence of complexes which stays short exact after $-{\otimes }_{R}M$.

The long exact sequence in its homology is the required sequence for the right-resolved Tor construction.

The balanced natural isomorphism identifies this with the stated balanced Tor functor, making the connecting maps independent of the chosen side.

The long exact Tor sequence contains ${\mathrm{Tor}}_{i+1}(N,P)\to {\mathrm{Tor}}_{i+1}(N,M)\to {\mathrm{Tor}}_i(N,K)\to {\mathrm{Tor}}_i(N,P)$.

Both outside terms vanish because $P$ is projective and $i\ge 1$.

Exactness therefore makes the middle arrow an isomorphism. Resolving a right module instead proves the other-variable form.

If $M$ is flat, tensor a full projective resolution $Q_{\bullet }\to N$ with $M$. Exactness of $-{\otimes }_{R}M$ preserves its augmented exactness, so $H_i(Q_{\bullet }{\otimes }_{R}M)=0$ for every $i>0$. The right-resolution definition and The balanced Tor bifunctor give ${\mathrm{Tor}}_1^R(N,M)=0$.

For right modules $A\to B\to C\to 0$, the universal property Universal property of the tensor product for balanced maps into abelian groups gives $C{\otimes }_{R}M\cong (B{\otimes }_{R}M)\mathrm{/}\mathrm{im}(A{\otimes }_{R}M)$: a balanced map on $B\times M$ kills that image exactly when it factors through $C\times M$. Thus tensoring is right exact over the arbitrary ring $R$. In particular $H_0(Q_{\bullet }{\otimes }_{R}M)\cong N{\otimes }_{R}M$, naturally under augmentation-preserving maps.

Conversely, let $K\hookrightarrow P$ be an inclusion of right modules. Apply the projective horseshoe construction to $0\to K\to P\to P\mathrm{/}K\to 0$, tensor its degreewise split resolution sequence with $M$, and take the long exact homology sequence. Its degree-zero boundary identifies the kernel of $K\otimes M\to P\otimes M$ with the image of ${\mathrm{Tor}}_1(P\mathrm{/}K,M)$.

The assumed vanishing makes every such tensor map injective; together with right exactness this gives exactness on all short exact sequences, hence flatness.

If $N$ is flat, tensor a full projective resolution of $M$ with $N$. Exactness of $N{\otimes }_R-$ preserves the augmented resolution, so its positive homology, and in particular ${\mathrm{Tor}}_1(N,M)$ by The balanced Tor bifunctor, vanishes.

If $A\to B\to C\to 0$ is exact in left modules, Universal property of the tensor product for balanced maps into abelian groups identifies $N{\otimes }_{R}C$ with $(N{\otimes }_{R}B)\mathrm{/}\mathrm{im}(N{\otimes }_{R}A)$: balanced maps annihilating the image are precisely those that descend to $N\times C$. Thus $N{\otimes }_R-$ is right exact over the arbitrary ring $R$. Applied to a resolution $P_1\to P_0\to X\to 0$, this gives $H_0(N{\otimes }_R{P}_{\bullet })\cong N{\otimes }_{R}X$. This is natural under comparison maps because augmentations commute with them. By The balanced Tor bifunctor, it is the natural identification ${\mathrm{Tor}}_0^R(N,X)\cong N{\otimes }_{R}X$.

Conversely, for an inclusion $K\hookrightarrow P$ of left modules, apply The long exact Tor sequence in the left-module variable to $0\to K\to P\to P\mathrm{/}K\to 0$. Exactness identifies the kernel of $N\otimes K\to N\otimes P$ with the image of ${\mathrm{Tor}}_1(N,P\mathrm{/}K)$.

Universal vanishing gives injectivity for every such inclusion, and right exactness supplies the rest; thus $N{\otimes }_R-$ is exact.

Its relevant segment is ${\mathrm{Tor}}_1(N,C)\stackrel{\mathrm{\partial }}{\to }N\otimes A\to N\otimes B\to N\otimes C\to 0$.

Exactness already holds at the last two positions by right exactness; the kernel at $N\otimes A$ is $\mathrm{im}\mathrm{\partial }$.

Thus injectivity of $N\otimes A\to N\otimes B$, and hence exactness of the whole tensor sequence, is equivalent to $\mathrm{\partial }=0$.

Tensoring this resolution with $M$ gives $0\to M\stackrel{n}{\to }M\to 0$ in degrees $1,0$.

Its degree-one homology is $\ker (n:M\to M)$.

That kernel is precisely the $n$-torsion subgroup, and the identification is natural in $M$.

The group ${\mathrm{Tor}}_1(\mathbb{Z}\mathrm{/}m,\mathbb{Z}\mathrm{/}n)$ is the kernel of multiplication by $m$ on $\mathbb{Z}\mathrm{/}n$.

Writing $g=\gcd (m,n)$, the congruence $mx\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ has exactly $g$ solutions modulo $n$.

Those solutions form the unique subgroup of the cyclic group $\mathbb{Z}\mathrm{/}n$ of order $g$, hence are isomorphic to $\mathbb{Z}\mathrm{/}g$.

Choose a projective resolution $0\to P_1\to P_0\to A\to 0$.

After tensoring with $B$, this complex has no terms in degrees $i\ge 2$.

Therefore its homology, which computes ${\mathrm{Tor}}_i^{\mathbb{Z}}(A,B)$, vanishes in every degree $i\ge 2$.

Every finitely generated subgroup of $M$ is free abelian, and $M$ is the filtered union of these free subgroups.

Tensoring a short exact sequence with a filtered union commutes with the filtered colimit; each free subgroup is flat, so each resulting sequence is exact.

Filtered colimits of abelian groups are exact, hence $-{\otimes }_{\mathbb{Z}}M$ is exact and $M$ is flat.

If $M$ is flat and $0\mathrm{\ne }r\in R$, tensor the injection $R\stackrel{\cdot r}{\to }R$ with $M$; multiplication by $r$ on $M$ is injective, so $M$ is torsion-free.

Conversely, each finitely generated submodule of a torsion-free module over a PID is free, and the module is their filtered union.

Free modules are flat and filtered colimits preserve exactness, so the same argument as for abelian groups makes $M$ flat.

Choose a projective resolution $P_{\bullet }\to M$. Because $R$ is commutative, it is a resolution by both left and right projective modules. The termwise symmetry maps $P_j{\otimes }_{R}N\to N{\otimes }_R{P}_j$, $p\otimes n\mapsto n\otimes p$, commute with the single chain differential.

Thus they give a natural chain isomorphism $P_{\bullet }{\otimes }_{R}N\cong N{\otimes }_R{P}_{\bullet }$. The first complex computes ${\mathrm{Tor}}_i^R(M,N)$ by resolving its right-module first variable; the second computes ${\mathrm{Tor}}_i^R(N,M)$ by resolving its left-module second variable.

Taking homology and using change-of-resolution coherence yields the claimed natural Tor symmetry.

If $F$ is flat, the Tor-one criterion gives ${\mathrm{Tor}}_1^R(N,F)=0$ for every right $N$, and dimension shifting in a projective resolution of $N$ gives ${\mathrm{Tor}}_i^R(N,F)=0$ for all $i>0$.

Suppose $0\to F_n\to \cdots \to F_0\to M\to 0$ is a flat resolution of length $n$. Break it into short exact sequences of successive kernels. The long exact Tor sequence and step 1.1 shift every ${\mathrm{Tor}}_i(N,M)$ with $i>n$ to a positive Tor group of $F_n$, hence to zero.

Conversely, take a projective resolution and put $K_0=M$ and $K_j=\ker (P_{j-1}\to P_{j-2})$ for $j\ge 1$, with $P_{-1}=M$. Repeated dimension shifting identifies ${\mathrm{Tor}}_1^R(N,K_n)$ with ${\mathrm{Tor}}_{n+1}^R(N,M)$; for $n=0$ this is the identity. The assumed vanishing makes $K_n$ flat by the Tor-one criterion. Truncating at $K_n$ gives a length-$n$ flat resolution, including the case $n=0$.

Every projective term in the resolution is flat.

Thus the same resolution is a flat resolution of $M$ of length at most $n$.

Taking the supremum of flat dimensions over all left, respectively right, modules gives $\mathrm{w.gl.dim}\le \mathrm{gl.dim}$ on each side.

For each $d\ge 0$, the criterion says that every left module has flat dimension at most $d$ if and only if ${\mathrm{Tor}}_i^R(N,M)=0$ for all typed pairs $(N,M)$ and every $i>d$. Thus the left weak dimension and the displayed Tor supremum have exactly the same finite upper bounds. Numbers in $\mathbb{N}\cup \{\mathrm{\infty }\}$ are determined by these upper bounds, proving their equality, including the infinite case.

Regard a left $R$-module $M$ as a right $R^{\mathrm{o}\mathrm{p}}$-module and a right $R$-module $N$ as a left $R^{\mathrm{o}\mathrm{p}}$-module, using The opposite ring $R^{\mathrm{o}\mathrm{p}}$. A supplied projective left resolution $P_{\bullet }\to M$ is also a projective right $R^{\mathrm{o}\mathrm{p}}$-resolution. The balanced tensor universal property Universal property of the tensor product for balanced maps into abelian groups gives chain isomorphisms $P_{\bullet }{\otimes }_{R^{\mathrm{o}\mathrm{p}}}N\to N{\otimes }_R{P}_{\bullet }$ by $p\otimes n\mapsto n\otimes p$. The balance relation is preserved since $(rp)\otimes n$ and $p\otimes (nr)$ both map to $nr\otimes p=n\otimes rp$. The inverse is the same flip, and the differentials commute because $N$ is in degree zero. By The balanced Tor bifunctor and the supplied-resolution balance theorem The left and right projective constructions of Tor are naturally isomorphic, this yields ${\mathrm{Tor}}_i^{R^{\mathrm{o}\mathrm{p}}}(M,N)\cong {\mathrm{Tor}}_i^R(N,M)$.

The tensor flip also identifies exactness of the tensor functors defining flatness, so a right $R$-module has the same flat dimension as its associated left $R^{\mathrm{o}\mathrm{p}}$-module. The given data on both hands supply the data needed for step 1.1 over $R^{\mathrm{o}\mathrm{p}}$. Its left weak dimension is therefore the right weak dimension of $R$, while step 1.2 identifies its Tor supremum with the one in step 1.1. This proves the asserted symmetry. In the zero ring every unital module is zero, every flat dimension is zero, and the Tor-degree set is empty, agreeing with the stated supremum convention.

A module may be resolved by the length-zero projective resolution concentrated in degree $0$.

Tensoring such a resolution has no positive homology, giving vanishing positive Tor.

Likewise an injective (or projective) resolution concentrated in degree $0$ has no positive cohomology, giving vanishing positive Ext.

Every abelian group has projective, hence flat, dimension at most one; therefore both dimensions are at most one.

The group ${\mathrm{Tor}}_1^{\mathbb{Z}}(\mathbb{Z}\mathrm{/}n,\mathbb{Z}\mathrm{/}n)\cong \mathbb{Z}\mathrm{/}n$ is nonzero.

The nonzero degree-one Tor forces weak global dimension at least one, while the known nonzero ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Z}\mathrm{/}n,\mathbb{Z})$ forces global dimension at least one.

The tensor construction requires its first input to be a right $R$-module so that $(xr)\otimes y=x\otimes (ry)$ is meaningful.

For two merely left modules, $xr$ is not part of the supplied structure, so the balancing relation is not typed.

Hence the displayed Tor expression is not even defined without additional bimodule or opposite-ring data, refuting the universal claim.

Resolving $M$ gives the left-resolved complex $\mathbb{Z}\mathrm{/}2\stackrel{3}{\to }\mathbb{Z}\mathrm{/}2$, whose differential is the identity. Resolving $N$ gives the right-resolved complex $\mathbb{Z}\mathrm{/}3\stackrel{2}{\to }\mathbb{Z}\mathrm{/}3$.

These complexes are not literally equal: even their degree-zero groups are $\mathbb{Z}\mathrm{/}2$ and $\mathbb{Z}\mathrm{/}3$. Nevertheless both have zero homology, as required because $\mathbb{Z}\mathrm{/}2\otimes \mathbb{Z}\mathrm{/}3=0$ and the positive Tor groups also vanish.

The tensor double-complex argument supplies a natural isomorphism only after passing to homology; this refutes equality by definition.

The group $\mathbb{Q}$ is torsion-free, hence flat over the PID $\mathbb{Z}$.

If $\mathbb{Q}$ were projective over $\mathbb{Z}$, it would be free; every nonzero free abelian group has a nonzero map to $\mathbb{Z}$, while ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=0$.

Thus $\mathbb{Q}$ is flat but not projective, so its projective dimension is not zero.

The module $\mathbb{Q}$ is flat because it is torsion-free over the PID $\mathbb{Z}$.

Flatness gives ${\mathrm{Tor}}_1^{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}\mathrm{/}2)=0$.

Neither $\mathbb{Q}$ nor $\mathbb{Z}\mathrm{/}2$ is projective as a $\mathbb{Z}$-module, so this is the required counterinstance.

The ordinary tensor product $M{\otimes }_{R}N$ already requires one factor to be right-handed.

Thus the supposed inputs of the symmetric expression are not generally a valid domain for Tor.

Commutativity supplies an identification of left and right actions; without it the asserted symmetry is ill-typed, not a theorem.

Tensoring $0\to I\to R\to R\mathrm{/}I\to 0$ with $k$ gives ${\mathrm{Tor}}_1^R(R\mathrm{/}I,k)\cong \ker (I{\otimes }_{R}k⟶k)$. The displayed multiplication map is zero because $I$ annihilates $k$.

Moreover $I{\otimes }_{R}k\cong I\mathrm{/}{I}^2$, and the residue classes of $x$ and $y$ form a $k$-basis of $I\mathrm{/}{I}^2$. Hence ${\mathrm{Tor}}_1^R(R\mathrm{/}I,k)\cong k^2$.

On the other hand, $\{m\in k:Im=0\}=k$. The Tor group has $k$-dimension two while the asserted $I$-torsion submodule has dimension one, so they are not equal (or even isomorphic).